Let f be a real valued continuous function defined on the positive real axis such that g(x)=∫_0^x t f(t) dt . If g (x^3 )=x^6+x^7 , then value of ∑_r=1^15 f (r^3 ) is :

Correct answer: (D) 310. g(x)=x^2+x^7 / 3 g^′(x)=2 x+7/3 x^4 / 3 f(x)=fracg^′(x)x f(x)=2+7/3 x^1 / 3 f (r^3 )=2+7/3 r ∑_r=1^15 (2+7/3 r )=2(15)+7/3 ((15(16))/2 ) =310

Q33 / 75 · JEE Main 2025 (28 Jan Shift 2) · Mathematics

Let $f$ be a real valued continuous function defined on the positive real axis such that $g(x)=\int_0^x \mathrm{t} f(\mathrm{t}) \mathrm{dt}$. If $\mathrm{g}\left(x^3\right)=x^6+x^7$, then value of $\sum_{r=1}^{15} f\left(\mathrm{r}^3\right)$ is :

A
$270$
B
$340$
C
$320$
Correct answer
D
$310$

Solution

$g(x)=x^2+x^{7 / 3}$ $\begin{aligned} & g^{\prime}(x)=2 x+\frac{7}{3} x^{4 / 3} \\ & f(x)=\frac{g^{\prime}(x)}{x} \\ & f(x)=2+\frac{7}{3} x^{1 / 3} \\ & f\left(r^3\right)=2+\frac{7}{3} r \\ & \sum_{r=1}^{15}\left(2+\frac{7}{3} r\right)=2(15)+\frac{7}{3}\left(\frac{15(16)}{2}\right)\end{aligned}$ $=310$

Hence, the correct answer is option: 'D'.

Let $f$ be a real valued continuous function defined on the positive real axis such that $g(x)=\int_0^x \mathrm{t} f(\mathrm{t}) \mathrm{dt}$. If $\mathrm{g}\left(x^3\right)=x^6+x^7$, then value of $\sum_{r=1}^{15} f\left(\mathrm{r}^3\right)$ is :

Solution

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